CEGEP CHAMPLAIN - ST. LAWRENCE
201-103-RE: Differential Calculus
Fall 2010 - Patrice Camiré
Derivatives of Exponential/Logarithmic Functions &
Logarithmic Differentiation
1. Use the rules of differentiation to differentiate the following functions.
(a) 𝑒2𝑥
(b) 𝑒𝑥
(f)
3
𝑒𝑥
𝑥+1
2
(j)
tan(𝑥)
(c) 4𝑒sin(𝑥)
√
(e) 𝑥𝑒3𝑥
(i) cos(𝑒
𝑥
(p) 𝑒𝑥 cos(2𝑥)
2𝑥
(l) 𝑒
(q) tan(𝑒𝑥 /𝑥)
(m) 6𝑥 63𝑥
)
3 𝑒𝑥
(o) sec(𝑒2𝑥 )
(k) 𝑥4 9𝑥
(g) 3
(h) sin(3𝑥)𝑒7𝑥
(d) 25𝑥
(n) 7𝑥
𝑒𝑥
𝑥−1
𝑥+1
(r) 𝑒 𝑥−1
2. Use the rules of differentiation to differentiate the following functions.
(a) 𝑥2 ln(𝑥)
(g) ln(cos 𝑥)
(b) ln(𝑥2 + 1)
(h) ln(sin 𝑥)
3
(c) ln(𝑥 )
(j)
𝑥(𝑥 − 1)3
(𝑥 + 1)4
(b) 𝑦 = 𝑥sin(𝑥)
(c) 𝑦 = 𝑥3 𝑒2𝑥 (𝑥2 + 1)7
𝑥4 𝑒5𝑥
(𝑥 − 1)3 (𝑥 + 1)2
√
(e) 𝑦 = 4𝑥 𝑒𝑥 + 1
√
√
3𝑥 + 1 3 𝑥 − 1
(f) 𝑦 =
𝑥2 + 8
(n) 𝑥 log2 (2𝑥 + 1)
ln(3𝑥 + 1)
2𝑥 + 1
(o) log4 (𝑥 cos(𝑥) tan(𝑥))
(k) log3 (𝑥)
3. Use logarithmic differentiation to find
(a) 𝑦 =
(m) log2 (𝑥) log9 (𝑥)
(i) ln(𝑥2𝑥 𝑒𝑥 )
(d) ln(3𝑥) − 4
(e) ln(5𝑥) − 8
(
)
𝑥−1
(f) ln
𝑥+1
(l) log8 (cos 𝑥)
(p)
1
log5 (𝑥)
𝑑𝑦
.
𝑑𝑥
(g) 𝑦 =
𝑥(𝑥 + 1)
(𝑥 + 2)(𝑥 + 3)
(h) 𝑦 = [sin(2𝑥)]cos(3𝑥)
√
𝑥𝑒4𝑥
(i) 𝑦 =
23𝑥 sin(𝑥)
(d) 𝑦 =
2
(j) 𝑦 = 𝑒𝑥 sin(𝑥) cos(𝑥)53𝑥
(k) 𝑦 = (𝑥 − 1)4 (𝑥 + 1)3 (𝑥 + 2)5
(l) 𝑦 = [ln(sec 𝑥)]𝑥
Answers
(h) [3 cos(3𝑥) + 7 sin(3𝑥)]𝑒7𝑥
1. (a) 2𝑒2𝑥
(b) 3𝑥2 𝑒
𝑥3
(i)
sin(𝑥)
−𝑒
√
(e) (3𝑥 + 1)𝑒3𝑥
(f)
2𝑥
(l) 2𝑥 𝑒
tan(𝑥)
ln(2)
(k)
(l)
(m)
(n)
(o)
(h) cot(𝑥)
𝑑𝑦
𝑑𝑥
(c)
𝑑𝑦
𝑑𝑥
(d)
𝑑𝑦
𝑑𝑥
(e)
𝑑𝑦
𝑑𝑥
(f)
𝑑𝑦
𝑑𝑥
(g)
𝑑𝑦
𝑑𝑥
(
1
3
4
+
−
𝑥 𝑥−1 𝑥+1
(p)
)
𝑒𝑥
𝑥
)(
𝑥−1
𝑥2
3(2𝑥 + 1) − 2(3𝑥 + 1) ln(3𝑥 + 1)
(3𝑥 + 1)(2𝑥 + 1)2
1
𝑥 ln(3)
− tan(𝑥)
ln(8)
log9 (𝑥) log2 (𝑥)
+
𝑥 ln(2)
𝑥 ln(9)
2𝑥
log2 (2𝑥 + 1) +
(2𝑥 + 1) ln(2)
[
]
1
1
1
− tan(𝑥) +
ln(4) 𝑥
sin(𝑥) cos(𝑥)
−1
𝑥 ln(5) [log5 (𝑥)]2
𝑥(𝑥 − 1)3
(𝑥 + 1)4
(
)
sin(𝑥)
= cos(𝑥) ln(𝑥) +
𝑥sin(𝑥)
𝑥
(
)
3
14𝑥
=
+2+ 2
𝑥3 𝑒2𝑥 (𝑥2 + 1)7
𝑥
𝑥 +1
(
)
4
3
2
𝑥4 𝑒5𝑥
=
+5−
−
𝑥
𝑥 − 1 𝑥 + 1 (𝑥 − 1)3 (𝑥 + 1)2
(
)
√
𝑒𝑥
𝑥
= ln(4) +
4
𝑒𝑥 + 1
2(𝑒𝑥 + 1)
√
(
)√
3
1
2𝑥
3𝑥 + 1 3 𝑥 − 1
=
+
−
2(3𝑥 + 1) 3(𝑥 − 1) 𝑥2 + 8
𝑥2 + 8
(
)
1
𝑥(𝑥 + 1)
1
1
1
=
+
−
−
𝑥 𝑥 + 1 𝑥 + 2 𝑥 + 3 (𝑥 + 2)(𝑥 + 3)
𝑑𝑦
=
𝑑𝑥
(
𝑥+1
(j)
1
𝑥
2
−2𝑒 𝑥−1
(r)
(𝑥 − 1)2
(m) 4 ln(6)6
2. (a) 𝑥(2 ln(𝑥) + 1)
2𝑥
(b) 2
𝑥 +1
3
(c)
𝑥
1
(d)
𝑥
1
(e)
𝑥
1
1
2
(f)
−
= 2
𝑥−1 𝑥+1
𝑥 −1
(g) − tan(𝑥)
(b)
(q) 𝑒 sec
4𝑥
(g) ln(3) sec (𝑥)3
(p) (cos(2𝑥)−2𝑥 sin(2𝑥))𝑒𝑥 cos(2𝑥)
𝑥
(k) 9 𝑥 (𝑥 ln(9) + 4)
2
3. (a)
2
𝑥 3
𝑥𝑒𝑥
(𝑥 + 1)2
(i) 1 + ln(2) +
(o) 2 sec(𝑒2𝑥 ) tan(𝑒2𝑥 )𝑒2𝑥
√
2 𝑥
(2𝑥2 − 2𝑥 − 1)𝑒𝑥
(j)
(𝑥 − 1)2
(d) 5 ln(2)25𝑥
ln(7)
√
𝑥 sin(𝑒 𝑥 )
(c) 4 cos(𝑥)𝑒
3 𝑒𝑥
(n) 𝑥2 (𝑥 + 3)𝑒𝑥 7𝑥
)
[
]
𝑑𝑦
= 2 cos(3𝑥) cot(2𝑥) − 3 sin(3𝑥) ln(sin(2𝑥)) [sin(2𝑥)]cos(3𝑥)
𝑑𝑥
(
)√
1 1
𝑥𝑒4𝑥
𝑑𝑦
=
+ 4 − 3 ln(2) − cot(𝑥)
(i)
𝑑𝑥
2 𝑥
23𝑥 sin(𝑥)
(h)
[
] 2
𝑑𝑦
= 2𝑥 + cot(𝑥) − tan(𝑥) + 3 ln(5) 𝑒𝑥 sin(𝑥) cos(𝑥)53𝑥
𝑑𝑥
)
(
𝑑𝑦
4
3
5
(𝑥 − 1)4 (𝑥 + 1)3 (𝑥 + 2)5
(k)
=
+
+
𝑑𝑥
𝑥−1 𝑥+1 𝑥+2
)
(
𝑑𝑦
𝑥 tan(𝑥)
[ln(sec 𝑥)]𝑥
(l)
= ln ln(sec 𝑥) +
𝑑𝑥
ln(sec 𝑥)
(j)